Question

The sum of the three consecutive number in Ap is 21 and their produt is the 231

Answer 1

let the consecutive terms in AP be a-d,a,a+d.
sum,
a-d+a+a+d = 21
3a=21
a=7
product,
(a-d)*(a)*(a+d)=231
(a-d)*(7)*(a+d)=231

(a-d)*(a+d)=33
putting value of a=7
a^2-d^2=33
49-d^2=33
-d^2=-16
d=4

Answer 2

Here is your solution


let,

Three numbers be a,a-d,a+d

so a+a-d+a+d = 21

a = 7

A/q
(a)(a-d)(a+d) = 231

(7)(7-d)(7+d) = 231

49- d2 = 231/7

d2 = 16

therefore d = +4 or - 4

so no(s) r : 3, 7 and 10 or 10, 7 and 3

hope it helps you